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\(\frac{1}{x} + \frac{4}{3x} - \frac{5}{6x} + 1 = 0\) \(\frac{6 + 8 - 5 + 6x}{6x} = 0\) \(\frac{9 + 6x}{6x} = 0 \implies 9 + 6x = 0\) \(6x = -9 \implies x = \frac{-3}{2}\)

(a) \(T_{n} = a + (n - 1)d\) (terms of an AP) Given a = -8; \(T_{7} = a + 6d = -8 + 6d\) \(T_{9} = a + 8d = -8 + 8d\) \(\frac{-8 + 6d}{-8 + 8d} = \frac{5}{8}\) \(5(-8 + 8d) = 8(-8 + 6d)\) \(-40 + 40d = -64 + 48d\) \(-40 + 64 = 48d - 40d \times 24 = 8d\) \( d = 3\) (b) Given r = 2 cm, m = 11.2g (i) \(V = \frac{4}{3} \pi r^{3}\) = \(\frac{4}{3} \times \frac{22}{7} \times 2^{3}\) = \(\frac{704}{21}\) = \(33.52 cm^{3} = 33.52 \times 10^{-6} m^{3}\) = \(3.352 \times 10^{-5} m^{3}\) (ii) \(Density = \frac{mass}{volume}\) = \(\frac{11.2}{33.52}\) = \(0.334 g/cm^{3}\) (iii) \(V = \frac{4}{3} \pi r^{3}\) = \(\frac{4}{3} \times \frac{22}{7} \times 3^{3}\) = \(\frac{2376}{21}\) = \(113.14 cm^{3}\) \(mass = density \times volume\) \(0.334 \times 113.14 = 37.789g\)

The statement that is not true about a rectangle is: II and III only The diagonals of a rectangle do not intersect at right angles. and does not have four lines of symmetry but two.

\(\bigtriangleup\)OEN = \(\bigtriangleup\)OGN OE = OG = r(radii) (r - 8)\(^2\) + 12\(^2\) = r\(^2\) r\(^2\) - 16r + 64 + 144 = r\(^2\) 16r = 64 + 144 16r = 64 + 144 = 208 \(\frac{208}{16}\) = 13cm FH = 2FO = 2OH = 2R = 2 x 13cm = 26cm

(a) SEE FIGURE A above., Longitude difference = 123^\circ - 67^\circ = 56^\circ (i) Distance between X and Y along the equator = \(\frac{56}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 0\) = \(\frac{56}{360} \times 2 \times \frac{22}{7} \times 6400\) = \(6257.78km \approxeq 6258km\) (ii) Latitude difference = 90^\circ Distance from the North pole = \(\frac{90}{360} \times 2 \times \frac{22}{7} \times 6400\) = \(10057.14km \approxeq 10057km\) (b) SEE FIGURE B above,., \(OP^{2} = PN^{2} + ON^{2}\) \(4^{2} + 3^{2} = 25\) \(OP = \sqrt{25} = 5cm\) \(OP = OR = 5cm\) \(\Delta NOR = scalene\) \(\frac{OR}{\sin 20} = \frac{3}{\sin < ORN}\) \(\frac{5}{\sin 20} = \frac{3}{\sin < ORN}\) \(\sin < ORN = \frac{3 \times \sin 20}{5}\) \(\sin < ORN = 0.2052\) \(< ORN = \sin^{-1} (0.2052)\) \(< ORN = 11.84^\circ \approxeq 12^\circ\)

Area of trapezium = \(\frac{1}{2} (a + b)h\) \(120 = \frac{1}{2} (5 + 7) \times h\) \(120 = 6h\) \(h = 20 cm\)

The given figure is a kite. Here are the properties of a kite: Two Pairs of Adjacent Sides: A kite has two pairs of adjacent sides that are equal in length. Diagonals: The diagonals of a kite are perpendicular to each other. One diagonal bisects the other. Symmetry: A kite has one line of symmetry, which runs along the longer diagonal. Angles: The angles between the unequal sides are equal. Specifically, the angles opposite the equal sides are also equal. Therefore, ∠PQR = ∠PSR is correct.

(a) See Figure A above. (i) r = 1m, h = 8m. \(V = \pi r^{2} h = \frac{22}{7} \times 1^{2} \times 8\) = \(\frac{176}{7}\) = \(25.14 m^{3}\) (ii) \(V = L \times B \times H\) \(\frac{176}{7} = 4 \times 12 \times h\) \(h = \frac{176}{7} \div 48\) \(h = \frac{11}{21}m\) = \(\frac{1100}{21} cm = 52.38 cm\) \(\approxeq 52 cm\) (nearest whole number) (b)(i) see figure B. \(3.3^{2} + 5.6^{2} = BD^{2}\) \(BD = \sqrt{3.3^{2} + 5.6^{2}}\) = \(\sqrt{10.89 + 31.36}\) = \(\sqrt{42.25} = 6.5 cm\) (ii) See Figure C If < BCD is right-angled, then Pythagoras rule holds true. \(BD^{2} = 5.2^{2} + 3.9^{2} = 27.04 + 15.21\) = \(42.25\) \(BD = \sqrt{42.25} = 6.5 cm\) Hence, \(\Delta BCD\) is right-angled.

\(\pi r^2\) = 38.5 \(\frac{22}{7} r^2\) = 38.5 r\(^2\) = \(\frac{38.5 \times 7}{22}\) r = \(\sqrt{12.25}\) = 3.5 diameter = 2r = 2 x 3.5 = 7cm

(a) Mean \(\bar{x} = \frac{\sum x}{n} = 8\) \(\frac{1 + 2 + x + 11 + y + 14}{6} = 8\) \(28 + x + y = 48 \implies x + y = 20 .... (1) \) The middle numbers are x and 11 therefore, \(\frac{x + 11}{2} = 9\) \(x + 11 = 18 \implies x = 18 - 11 = 7\) Putting x = 7 in (1), \(7 + y = 20 \implies y = 20 - 7 = 13\) (x, y) = (7, 13). (b) \(\frac{LM}{LP} = \frac{3}{4}\) \(\frac{\text{Area of LMN}}{\text{Area of LPQ}} = \frac{3^{2}}{4^{2}}\) Area of \(\Delta\) LPQ = \(\frac{16}{9} \times 18 = 32 cm^{2}\) \(\therefore \text{Area of quadrilateral MPQN} = 32 - 18 = 14 cm^{2}\)

Volume of pyramid with square base = \(\frac{\text{base area} \times height}{3}\) \(80 = \frac{\text{base area} \times 15}{3}\) \(80 = 5 \times \text{base area}\) \(\text{Base area} = 16 cm^2\)

Total volume = (5 x 5) x 5 + (5 x 5) x 2 = 125 + 50 = 175 cm\(^3\)

< T = < S = 50^\circ (OS = OT) < SOT = 180^\circ - 2(50^\circ) = 80^\circ < ROP = 80^\circ (vertically opposite angle) \(\therefore\) < OPQ = 180^\circ - 80^\circ = 100^\circ (adjacent angles)

Sum of interior angle in a hexagon = (6 - 2) x 180^\circ = 720^\circ \(\therefore\) 107^\circ + 2x^\circ + 150^\circ + 95^\circ + (2x - 15)^\circ + 123^\circ = 720^\circ 460 + 4x = 720 \(\implies\) 4x = 720 - 460 4x = 260^\circ \(\implies\) x = 65^\circ

(a) \(\cos x = 0.7431\) \(x = \cos^{-1} (0.7431)\) \(x = 42^\circ\) (i) \(2 \sin x = 2 \sin 42\) = \(2 \times 0.6692\) = \(1.3384\) (ii) \(\tan \frac{x}{2} = \tan \frac{42}{2}\) = \(\tan 21^\circ\) = \(0.3839\) (b) Sum of the interior angles of a polygon = \((2n - 4) \times 90^\circ\) For a pentagon, n = 5 \((2(5) - 4) \times 90^\circ = 6 \times 90^\circ\) = \(540^\circ\) Ratio of sides = 2:3:4:4:5 Total = 2 + 3 + 4 + 4 + 5 = 18 Largest angle = \(\frac{5}{18} \times 540^\circ = 150^\circ\)